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alexdok [17]
3 years ago
5

Help, please! instructions in the picture:

Mathematics
2 answers:
Paraphin [41]3 years ago
7 0
4 hours is $50, 40 hours is $500, 160 hours is $2000, 200 hours is $2500.

i’m pretty sure this is right because she makes $12.5 an hour.
Gelneren [198K]3 years ago
3 0

Answer:

$ 500 ~40 horus

$ 2000 ~160hours

$ 2500 ~200hours

Step-by-step explanation:

Assume that the rate of earnings is constant, then one can manipulate the first ratio to get to the succeeding ratios. A ratio is a way of expressing one value compared to another when the two values have different units. In this case, the ratio is: (hours : earnings)

hours : earnings

4 : 50

First, one is asked to find how much is earned in (40) hours, since the rate is constant, all one has to do is multiply the hours by (10) to get to (40),

40 : 500

Next, one is asked to find how much is earned in (160) hours, now multiply the previous rate by (4), to get to 160,

160: 2000

Finally, one is asked to find out how much is earned in (200) hours, multiply the rate which gives one how much is earned in (40) hours by (5),

200 : 2500

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PLSSS HELP ME OUT WITH THIS QUESTION, PLS PLS PLS AND PLS EXPLAIN!!!
Marina CMI [18]

Answer:

C. 15

Step-by-step explanation:

1. First, let's solve the system of equations; -m - n = 5, m - n = -3

Solve -m - n = 5 for m.

  • -m -n = 5
  • -m-n+n=5+n
  • -m = n + 5
  • \frac{-m}{-1} = \frac{n+5}{-1}
  • m = -n - 5

Substitute -n - 5 for m in m - n = -3.

  • m - n = -3
  • -n - 5 - n = -3
  • -2n-5=-3
  • -2n-5+5=-3+5
  • -2n = 2
  • \frac{-2n}{-2} = \frac{2}{-2}
  • n = -1  

Substitute -1 for n in m = -n - 5.

  • m = -n - 5
  • m = -(-1) - 5
  • m = 1 - 5
  • m = -4

2. Okay, now that we know m = -4 and n = -1, let's solve for m^2 -n^2.

  • m^2-n^2
  • (-4)^2-(-1)^2
  • 16 - (-1)^2
  • 16 - 1
  • 15

Therefore, the answer is C. 15!

4 0
3 years ago
Read 2 more answers
Are 1/2=2/3=3/4=4/5 all equal to each other
melisa1 [442]

Answer:

nooo lol

Step-by-step explanation:

1/2 is NOT gonna be equal to 4/5 remember, if u have 4 slices of pizza thats only equal to 1/2 but if u have 7 slices u have 4/5

8 0
3 years ago
Read 2 more answers
1/4 -
marissa [1.9K]
Definitely socratic it’ll help
8 0
3 years ago
What is. 5(x-1)+3=3x-2(3-2x) <br><br> I give Brainliest! Please show your work
garik1379 [7]

Answer:

x=2

Step-by-step explanation:

Expand 5(x-1)+3: 5x-2

5x-2=7x-6

<em>add 2 both sides and simplify</em>

5x=7x-4

<em>subtract 7x from both sides</em>

5x=7x-4-7

___-7____

<em>simplify -2x=4x divide both sides</em> by 2, -2x/-2=-4/-2

x=2

3 0
3 years ago
Read 2 more answers
Which pair of funtions is not a pair of inverse functions? please help!!
antiseptic1488 [7]

Answer:

f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}

Step-by-step explanation:

we know that

To find the inverse of a function, exchange variables x for y and y for x. Then clear the y-variable to get the inverse function.

we will proceed to verify each case to determine the solution of the problem

<u>case A)</u> f(x)=\frac{x+1}{6} , g(x)=6x-1

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

x=\frac{y+1}{6}

Isolate the variable y

6x=y+1

y=6x-1

Let

f^{-1}(x)=y

f^{-1}(x)=6x-1

therefore

f(x) and g(x) are inverse functions

<u>case B)</u> f(x)=\frac{x-4}{19} , g(x)=19x+4

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

x=\frac{y-4}{19}

Isolate the variable y

19x=y-4

y=19x+4

Let

f^{-1}(x)=y

f^{-1}(x)=19x+4

therefore

f(x) and g(x) are inverse functions

<u>case C)</u> f(x)=x^{5}, g(x)=\sqrt[5]{x}

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

x=y^{5}

Isolate the variable y

fifth root both members

y=\sqrt[5]{x}

Let

f^{-1}(x)=y

f^{-1}(x)=\sqrt[5]{x}

therefore

f(x) and g(x) are inverse functions

<u>case D)</u> f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

x=\frac{y}{y+20}

Isolate the variable y

x(y+20)=y

xy+20x=y

y-xy=20x

y(1-x)=20x

y=20x/(1-x)

Let

f^{-1}(x)=y

f^{-1}(x)=20x/(1-x)

\frac{20x}{1-x}\neq \frac{20x}{x-1}

therefore

f(x) and g(x) is not a pair of inverse functions

7 0
3 years ago
Read 2 more answers
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